Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Solution to Exercise 9

a) If the number of (occupied!) edge states equals N, the Fermi energy E_{F} must be somewhere between (N − 1/2)h̄ω_{c} and (N + 1/2)h̄ω_{c}. If N is much larger than 1, we may ignore the zero point energy, and

N ≃ \frac{E_{F}}{h̄ω_{c}} .

b) According to Büttiker and Landauer,

I_{α} = ∑_{β≠α} G_{αβ} (V_{α} − V_{β}) ,

with conductances

G_{αβ} = \frac{2e^{2}}{h} T_{αβ}.

Here, T_{αβ} is the ”transmissivity” from terminal β to terminal α. Let us e.g. assume that the magnetic field B is pointing out of the plane. Then, electrons will be deflected to the left, so that edge states propagating from left to right are created near the upper edge of the channel, and edge states propagating from right to left are created near the lower edge of the channel. All the N edge states entering the system at the upper edge in terminal 1 will follow the upper edge and be transmitted into terminal 2. Hence, T_{21} = N. Only n of the N states that enter the system at the right edge of terminal 2 are transmitted through the constriction and end up in terminal 4. The remaining N − n states continue along the lower edge and finally end up in terminal 1. Hence, T_{42} = n and T_{12} = N − n. Similar arguments further yield T_{34} = N, T_{13} = n, and T_{43} = N − n. The remaining six T_{αβ} are all zero. The figure below illustrates this, for N = 4 and n = 2.

Note that away from the narrow constriction, electron transport takes place via edge states, where the number of states N are determined by the magnetic field B. However, inside the constriction, electrons propagate via transverse modes, where the number of states n are determined by the geometry, i.e., the width w of the constriction.

Based on these considerations, we obtain the following equations relating currents and potentials:

I_{1} = \frac{2e^{2}}{h} \{(N − n) (V_{1} − V_{2}) + n (V_{1} − V_{3})\}
= \frac{2e^{2}}{h} \{NV_{1} − (N − n) V_{1} − nV_{3}\}
I_{2} = \frac{2e^{2}}{h} \{N (V_{2} − V_{1})\} = \frac{2e^{2}}{h} \{−NV_{1} + NV_{2}\}
I_{3} = \frac{2e^{2}}{h} \{N (V_{3} − V_{4})\} = \frac{2e^{2}}{h} \{NV_{3} − NV_{4}\}
I_{4} = \frac{2e^{2}}{h} \{n (V_{4} − V_{2}) + (N − n) (V_{4} − V_{3})\}
= \frac{2e^{2}}{h} \{−nV_{2} − (N − n) V_{3} + NV_{4}\}

As suggested, we choose V_{3} = 0. The longitudinal 4–terminal resistance R_{L} is found by measuring the voltage V_{2} − V_{4} when a current I_{1} = −I_{3} is established (with an applied voltage V_{1} − V_{3} = V_{1}). In other words, terminals 2 and 4 are voltage probes, with zero net current, i.e., I_{2} = I_{4} = 0. Zero current in terminal 2 yields V_{2} = V_{1}, and zero current in terminal 4 yields V_{4} = nV_{2}/N = nV_{1}/N. Hence,

R_{L} = R_{13,24} = \frac{V_{2} − V_{4}}{I_{1}} = \frac{h}{2e^{2}} \frac{V_{1} (1 − n/N)}{V_{1} (N − N + n)} = \frac{h}{2e^{2}} \frac{N − n}{Nn} = \frac{h}{2e^{2}} (\frac{1}{n} − \frac{1}{N} ) .

c) As already discussed above, the number of transmitted modes n is determined by the constriction width w for a given value of the Fermi energy E_{F}. As long as the magnetic length l_{B} is larger then w, n will be independent of the magnetic field. On the other hand, the number of edge states N in the wide region equals the number of occupied Landau levels, N ≃ E_{F} /h̄ω_{c} = E_{F} m/e|B|. Hence,

R_{L}(B) = \frac{h}{2e^{2}} (\frac{1}{n} − \frac{e|B|}{mE_{F}} ) .

This result predicts a maximum value of R_{L} for B = 0, and a linear dependence on |B|, in agreement with the experimental results reported in figure 50 in Beenakker/van Houten. (Again: For ”moderate” values of B!)

d) The 2–terminal resistance is

R_{2t} = R_{13,13} = \frac{V_{1} − V_{3}}{I_{1}} = \frac{h}{2e^{2}} \frac{V_{1}}{nV_{1}} = \frac{h}{2e^{2}} \frac{1}{n}.

This is as expected: There are n ”open channels” through the constriction between terminals 1 and 3.